Chapter 7 and 8 Suggested Problems

 

Chapter 7

1. For each of the following sampling plans, indicate why the target population and the sampled population are no the same.

a) In order to determine the opinions and attitudes of customers who regularly shop at a particular mall, a surveyor stands outside a large department store in the mall and randomly selects people to participate in the survey.

People who enter a particular store in the mall may not shop at all stores and therefore may not represent the average mall shopper.  Some people go to the mall only for one store.  To get opinions of people who regularly go to the mall. You must collect information/data over a long period of time such as several days over a periods of several months.

b) A library wants to estimate the proportion of its books that has been damaged. The librarians decide to select one book per shelf as a sample by measuring 12 inches from the left edge of each shelf and selecting the book in that location.

Not a random sample.

c) Political surveyors visit 200 residences during one afternoon at ask eligible voters present in the house at the time whom they intend to vote for.

People at home during the time period in which the survey was conducted may not represent the average voter and is only a small portion of the voting population.

 

2. A statistician would like to conduct a survey to ask people their views on a proposed new shopping mall in their community. According to the latest census, there are 800 households in the community. The statistician has numbered each household (from 1 to 800) and she would like to randomly select 25 of these households to participate in the study. Use Minitab to generate the sample.

Minitab Menu commands:

1. Click Calc, random data and integer

2. Type the number of random variables to be generated (50)

3. Hit tab and type the column where the numbers are to be stored

4. Hit tab and type the minimum value (1)

5. Hit tab and type the Maximum value (800)

6. Click O.K.

7. Print the Column of stored numbers

 

3. The operations manager of a large plant with four departments wants to estimate the person-hours lost per month due to accidents. Describe a sampling plan that would be suitable for estimating the plant-wide loss and for comparing departments.

Select stratified random samples where the samples are the four departments.  Then conduct random sampling in each department.

 

4. Work Exercise 7.26 from your Barnes text. The service times for customers at a supermarket checkout are independent random variables with a mean of 2 minutes and a variance of 1 minute.  What is the probability that 50 customers can be serviced in less than 2 hours?

Let Xi denote the service time for the ith customer.  We want to find:

Since the sample size is large, the central limit theorem tells us that this distribution is approximately normal with:

 

5. Work Exercise 7.27 from your Barnes text.  Suppose that SAT test scores from all high school seniors in the “Crabgrass State” have a mean of 1000 and a variance of 1600.  If a specific high school class of n = 36 students has an average score of x-bar = 985, is there any evidence to suggest that this high school class’ score is inferior to the state average?

Let X-bar denote the sample mean of a sample of n = 36 scores from a population with m = 1000 and s = 40.  We want to compute:

Since the probability is so small, it is unlikely that this school class can be regarded as a random sample from a population with a mean of 1000 and a standard deviation of 40.  There is evidence to suggest that the average score for this school is lower than the mean population.

 

6. Exercise 7.18 from your Barnes text A sample of size n = 45 is to be drawn from a gamma distribution with l = 2 and r = 6.  What is (P x-bar > 6)?

Using the Barnes software:

(P x-bar > 6) = .0203

m = r/l = 6/2 = 3 = E(X) = mx

s2 = r/l2 = 6/4 = 3/2 = 1.5 = 1.225

 

7. Exercise 7.22 from your Barnes text. A sample of size 36 is to be drawn from a Poisson distribution with parameter l = 4.  What is (P x-bar > 4.3)?

We can invoke the Central Limit Theorem since the sample size is > 30. Thus x-bar is distributed as a Normal distribution.

 

8. Exercise 7.19 from your Barnes text.  The weights of cartons of conical bearings used in the assembly of gravity driven conveyor systems is governed by a distribution with a mean of 25 pounds and a standard deviation of 2.5 pounds.  The motorized cart used to transport the bearings from the loading dock to the assembly point is rated at a maximum of 890 pounds.  If efficiency in delivering the cartons is important, how many cartons should be loaded on the cart for each trip?

We need an “n” such that P(T>890) is very small.

We will use n = 34 since 850 plus 2 standard deviations will have a 95% chance of being under 890 pounds:

850 + 14.57 + 14.57 = 879.14

 

Chapter 8

Note for Data_ch8: Go through hyperlink to data file Data_ch8. Save the file as "html" to your floppy disc. Open the file in Excel 97. Save the file as an Excel 97 file. Copy the data for each problem into Minitab and calculate the answers.

 

6. In a survey conducted to determine the cost of vacations, 63 individuals were randomly sampled. Each person was asked to compute the cost of his or her most recent vacation. The data is stored in Data_ch8. Assuming the standard deviation is $400, estimate with 95% confidence the average cost of vacations.

Using Minitab:

1.      Type or import the data into one column

2.      Click Stat, Basic Statistics, and 1-Sample z

3.      Type variable name (Time or C1)

4.      Use the cursor to select Confidence interval

5.      Hit tab and type the confidence level (95)

6.      Hit tab and type the value of the population standard deviation (sigma = 8.0)

7.      Click O.K.

One-Sample Z: Cost

 

The assumed sigma = 400

 

Variable          N      Mean     StDev   SE Mean         95.0% CI   

Cost                64    1350.0     347.2      50.0          (1252.0,  1448.0)

 

7. As part of a project to develop better lawn fertilizers, a research chemist wanted to determine the mean weekly growth rate of Kentucky Bluegrass, a common type of grass. A sample of 250 blades of grass was measured, and the amount of growth in one week was recorded. These produced an x-bar of .89.

  1. Assuming that weekly growth is normally distributed with a standard deviation of .10 inches, estimate with 99% confidence the mean weekly growth of Kentucky Bluegrass.
  2. Briefly interpret what the interval estimate tells you about the growth of Kentucky bluegrass.

Z a/2 = z .005 = 2.575

n = 250

x-bar = .89

(.874, .906)

 

8. A medical statistician wants to estimate the average weight loss of people who are on a new diet plan. In a preliminary study, he found that the smallest weight loss was three pounds and the largest weight loss was 39 pounds. How large a sample should he take to estimate the mean weight loss to within two pounds, with 90% confidence?

The confidence level is 1 - a = .90

Z a/2 = z .05 = 1.645

The error bound is B = 2 pound

The range = 39 – 3 = 36 pounds

We approximate s as: s ~ range/4 = 36/4 = 9 pounds

Now we can solve for n:

Which rounds up to 55.

 

9. Exercise 8.5 from your Barnes text.  Seventeen bushings yield s = .0004 inch in the measurement of their internal diameters. Set a 90% confidence interval on the true standard deviation of the bushings’ internal diameters.

What do we know?

n = 17

s = .0004 inch

 

10. Exercise 8.34 from your Barnes text.  A nuclear power plant used up plutonium rods in the following amounts: 26.3, 32.2, 19.7, 21.5, 27.1, and 17.2 tons per month.  If a report provided by the supervisor stated that the mean number of rods used per month has an equal-tail confidence interval of (20.4, 27.96), what level of confidence is associated with the interval?

x-bar = 26.3 + 32.2 + 19.7 + 21.5 + 27.1 + 17.2 = 144

since n < 30, but we assume normality, then we use:

now try different values of alpha to make this equation true.  See table A.5:

t 5,.10 = 1.476

t 5,.05 = 2.015

Now interpolate:

 

Thus alpha/2 = .075 and alpha = .15